integral from 0 to 7 of t^3-9t^2+20t

Lösung

\int_{0}^{7} t^{3} - 9 t^{2} + 20 t d t

\frac{245}{4}

Dezimale

61.25

Schritte zur Lösung

\int_{0}^{7} t^{3} - 9 t^{2} + 20 t d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{7} t^{3} d t - \int_{0}^{7} 9 t^{2} d t + \int_{0}^{7} 20 t d t

\int_{0}^{7} t^{3} d t = \frac{2401}{4}

\int_{0}^{7} 9 t^{2} d t = 1029

\int_{0}^{7} 20 t d t = 490

= \frac{2401}{4} - 1029 + 490

Vereinfache

\frac{2401}{4} - 1029 + 490 : \frac{245}{4}

= \frac{245}{4}

\int_{0}^{3} x^{2} cos (x) d x

\int_{0}^{1} x^{3} (x - \frac{1}{2}) d x

\int_{0}^{4} x (x - 2) (x - 4) d x

\int_{0}^{2} 2 t \cdot e^{t^{2}} d t

\int_{0}^{4} \frac{x}{x + 9} d x